Let ABC be a triangle. Consider four points p _ 1 , p _ 2 , p _ 3 , p _ 4 on the side AB, five points p_ 5 , p_ 6 , p_ 7 , p_ 8 , p_ 9 on the side B C, and four points p_ 10 , p_ 11 , p_ 12 , p_ 13 on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentago

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Let ABC be a triangle. Consider four points p _ 1 , p _ 2 , p _ 3 , p _ 4 on the side AB, five points p_ 5 , p_ 6 , p_ 7 , p_ 8 , p_ 9 on the side B C, and four points p_ 10 , p_ 11 , p_ 12 , p_ 13 on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p _ 1 , p _ 2 , , p _ 13 , is \_\_\_\_

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Correct answer: A. A

Points: 4 on AB, 5 on BC, 4 on AC. For a valid pentagon, no 3 vertices can be collinear, so at most 2 points from each side. Only valid distribution (a, b, c) with a+b+c = 5 and each 2 is permutations of (1, 2, 2): (1,2,2): 4 1 5 2 4 2 = 4 10 6 = 240 (2,1,2): 4 2 5 1 4 2 = 6 5 6 = 180 (2,2,1): 4 2 5 2 4 1 = 6 10 4 = 240 Total = 240 + 180 + 240 = 660.

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